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Step-by-Step Solution
Step 1: Recall the relation between de Broglie wavelength and kinetic energy
The de Broglie wavelength $ \lambda $ of a particle is related to its momentum $ p $ by:
$$ \lambda = \frac{h}{p}. $$
For a particle of mass $ m $ and kinetic energy $ E = \frac{1}{2} m v^2 $, the momentum is $ p = mv = \sqrt{2mE} $. Hence,
$$ \lambda = \frac{h}{\sqrt{2mE}}. $$
This shows that $ \lambda \propto \frac{1}{\sqrt{E}} $.
Step 2: Relate the change in wavelength to the change in energy
We are told the de Broglie wavelength must reduce to $75\%$ of the initial value, i.e.,
$$ \lambda_2 = 0.75 \, \lambda_1. $$
Since $ \lambda \propto \frac{1}{\sqrt{E}} $, we can write:
$$ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{E_1}{E_2}}. $$
Substitute $ \lambda_2 = 0.75 \,\lambda_1 $:
$$ 0.75 = \sqrt{\frac{E_1}{E_2}}. $$
Step 3: Express the new energy in terms of the initial energy
Squaring both sides gives:
$$ 0.75^2 = \frac{E_1}{E_2} = 0.5625. $$
Thus,
$$ \frac{E_1}{E_2} = \left( \frac{3}{4} \right)^2 = \frac{9}{16}, $$
which rearranges to
$$ E_2 = \frac{16}{9} E_1. $$
Since the initial kinetic energy is $ E_1 = E $, we find
$$ E_2 = \frac{16}{9} E. $$
Step 4: Calculate the extra energy required
The additional energy needed is the difference between the new energy $ E_2 $ and the original energy $ E_1 $:
$$ \text{Extra energy} = E_2 - E = \frac{16}{9}E - E = \left(\frac{16}{9} - 1\right)E = \frac{7}{9}E. $$
